Rings
Table of Contents

Rings

Recall from the Groups page that a group is a set $G$ with a (closed) binary operation $\cdot$ denoted $(G, \cdot)$ such that:

  • 1. The operation $\cdot$ is associative, that is, for all $a, b, c \in G$ we have that $a \cdot (b \cdot c) = (a \cdot b) \cdot c$.
  • 2. There exists an element $e \in G$ such that for all $a \in G$ we have that $a \cdot e = a$ and $e \cdot a = a$ to which we call $e$ the unique identity of $\cdot$ in $G$.
  • 3. For each $a \in S$ there exists an $a^{-1} \in G$ such that $a \cdot a^{-1} = e$ and $a^{-1} \cdot a = e$ to which we call $a^{-1}$ the inverse of $a$.

We will now describe a new type of structure called a ring.

Definition: If $+$ and $\cdot$ are (closed) binary operations on a (nonempty) set $R$, then $R$ is called a Ring under $+$ and $\cdot$ denoted $(R, +, \cdot)$ if $R$ under $+$ and $\cdot$ satisfies the following properties:
1. For all $a, b, c \in R$, $a + (b + c) = (a + b) + c$ (Associativity of elements in $R$ under $+$).
2. There exists an $0 \in R$ such that for all $a \in R$ we have that $a + 0 = a$ and $0 + a = a$ (The existence of an identity element $0$ of $R$ under $+$).
3. For all $a \in R$ there exists a $-a \in R$ such that $a + (-a) = 0$ and $(-a) + a = 0$ (The existence of inverses for each element in $R$ under $+$).
4. For all $a, b \in R$ we have that $a + b = b + a$ (Commutativity of elements in $R$ under $+$).
5. For all $a, b, c \in R$, $a \cdot (b \cdot c) = (a \cdot b) \cdot c$ (Associativity of elements in $R$ under $\cdot$).
6. There exists a $1 \in R$ such that for all $a \in R$ we have hat $a \cdot 1 = a$ and $1 \cdot a = a$ (The existence of an identity element $1$ of $R$ under $\cdot$).
7. For all $a, b, c \in R$ we have that $a \cdot (b + c) = (a \cdot b) + (b \cdot c)$ and $(a + b) \cdot c = (a \cdot c) + (b \cdot c)$ (Distributivity of $\cdot$ over $+$).

The operation $+$ is commonly referred to as addition while the operation $\cdot$ is commonly referred to as multiplication. In the definition above, instead of using $e$, we use $0$ to denote the identity element of the operation $+$ commonly referred to as the Additive Identity, and we use $1$ to denote the identity element of the operation $\cdot$ commonly referred to as the Multiplicative Identity.

Sometimes the definition of a ring omits axiom (6). In this case, the term "Ring with Unit" is used to described our definition above, and the term "Ring Without Unit" describes axioms (1)-(5) and (7) above.

Example 1

One example of a ring in the set of rational numbers $\mathbb{Q}$ with the operations $+$ of standard addition and $\cdot$ of standard multiplication. Let's verify this.

It's easy to verify that $\mathbb{Q}$ is closed under $+$ and that $+$ is associative. Furthermore, the additive identity is $0 \in \mathbb{Q}$. For each $x \in \mathbb{Q}$ we have that $x = \frac{a}{b}$ for $a, b \in \mathbb{Z}$ and $b \neq 0$ and the additive inverse of $x$ is $-x = - \frac{a}{b} \in \mathbb{Q}$ since $\displaystyle{x + (-x) = \frac{a}{b} + \left ( -\frac{a}{b} \right ) = 0}$. Furthermore, it is obvious that the addition of rational numbers is commutative.

If $x, y \in \mathbb{Q}$ where $\displaystyle{x = \frac{a}{b}}$ and $\displaystyle{y = \frac{c}{d}}$ with $a, b, c, d \in \mathbb{Z}$ and $b, d \neq 0$ then the product of $a$ and $b$ is closed under $\cdot$ since $ac, bd \in \mathbb{Z}$ and $bd \neq 0$:

(1)
\begin{align} \quad x \cdot y = \frac{a}{b} \cdot \frac{c}{d} = \frac{ac}{bd} \in \mathbb{Q} \end{align}

It can be easily verified that $\cdot$ is an associated operation for addition of rational numbers.

The multiplicative identity is the rational number $\displaystyle{1 = \frac{1}{1}}$

Lastly, it's not hard to show that the distributivity property holds. Let $x = \frac{a}{b}, y = \frac{c}{d}, z=\frac{e}{f} \in \mathbb{Q}$. For left distributivity we have:

(2)
\begin{align} \quad x \cdot (y + z) = \frac{a}{b} \cdot \left ( \frac{c}{d} + \frac{e}{f} \right ) = \frac{ac}{bd} + \frac{ae}{bf} = x *y + x \cdot z \end{align}

For right distributivity we have:

(3)
\begin{align} \quad (x + y) \cdot z = \left ( \frac{a}{b} + \frac{c}{d} \right ) \cdot \frac{e}{f} = \frac{ae}{bf} + \frac{ce}{df} = x*z + y*z \end{align}

Therefore $(\mathbb{Q}, +, \cdot)$ is a ring.

Example 2

It is not hard to check that $(\mathbb{C}, + \cdot)$, $(\mathbb{R}, +, \cdot)$, and $(\mathbb{Z}, +, \cdot)$ are also rings.

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